THE NUMBER N!!!!!!!!!!!!
Level
pending
N is a positive integer such that it is the product of 3 distinct primes. Find all such natural numbers such that the sum of all its composite divisors is 2N+1.
The answer is 1.
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1 solution
Let N = abc where a,b and c are prime numbers. It is given ab + bc + ca + abc = 2abc + 1 or ab + bc +ca = abc + 1. Now let's suppose that none of a,b, and c is even prime i.e. 2. But this will generate a contradiction that L.H.S becomes odd and R.H.S becomes even. So one of a,b,c is 2; without the loss of generality let's take a = 2. After substitution we will get this equation, 2b + 2c = bc + 1 or b = (2c - 1)÷(c-2) or b = 2 + 3÷(c-2). This implies that c-2 should divide 3 so c = 3 and 5; b=5 and 3 for c = 3 and 5 respectively. Hence N = 2×3×5 = 30. Only one solution.